# From finitely additive to countably additive

Given a finitely additive measure $\mu : \mathcal{B} \rightarrow [0,\infty]$. If we want to show that it is also countably additive, finite additivity implies $$\mu(E) \geq \sum_{n=1}^\infty \mu(E_n)$$ where $\cup E_n = E$ with $E_n$ disjoint automatically.

For the inequality in the other direction, clearly countably subadditive is sufficient. I was wondering if there are weaker conditions that would also be sufficient. (Just take $\mathcal{B}$ to be the Borel $\sigma$-algebra on $\mathbb{R}$)

I know for fact that if $\mu$ is a Radon measure would also be enough (maybe $\sigma$-finite with some regularity properties).

There are several easy criterions other than those you have mentioned that, together with finite additivity, imply countable additivity (but they all end up implying countable subadditivity in the end). Here are two examples:

1. If $\mu$ is finitely additive and continuous from below (that is, if $A_1\subset A_2\subset\cdots$ and $A=\bigcup_nA_n$, then $\mu(A_n)\to A$), then $\mu$ is countably additive: Take $\{A_n\}$ disjoint, and define $B_n=A_1\cup\cdots\cup A_n$. Then, $B_1\subset B_2\subset\cdots$ and $\bigcup_nB_n=\bigcup_nA_n$, and thus, finite additivity implies that $$\lim_{n\to\infty}\sum_{i=1}^n\mu(A_i)=\lim_{n\to\infty}\mu(B_n)=\mu\left(\bigcup_{n}A_n\right).$$

2. If $\mu$ is finitely additive and continuous from above at $\varnothing$ (that is, if $A_1\supset A_2\supset\cdots$ and $\bigcap_nA_n=\varnothing$, then $\mu(A_n)\to0$), then $\mu$ is countably additive: Take $\{A_n\}$ disjoint, and define $B_n=A_1\cup\cdots\cup A_n$. Then, $\bigcup_nB_n=\bigcup_nA_n$, and for every $m$, we have by finite additivity that $$\mu\left(\bigcup_nA_n\right)=\mu\left(B_m\cup\left(\bigcup_nA_n\right)\setminus B_m\right)=\mu\left(B_m\right)+\mu\left(\left(\bigcup_nA_n\right)\setminus B_m\right).$$ Now, we see that $\left(\bigcup_nA_n\right)\setminus B_1\supset\left(\bigcup_nA_n\right)\setminus B_2\supset\cdots$, and that $\bigcap_m\left(\left(\bigcup_nA_n\right)\setminus B_m\right)=\varnothing$. Thus, $$\mu\left(\bigcup_nA_n\right)=\lim_{m\to\infty}\left[\mu\left(B_m\right)+\mu\left(\left(\bigcup_nA_n\right)\setminus B_m\right)\right]=\lim_{m\to\infty}\mu(B_m)+0.$$ Since $\mu(B_m)=\sum_{i=1}^m\mu(A_i)$ for each $m$, this implies countable additivity.

user78270's answer is good, but leaves out some details. The following is from Measure Theory by F. R. Halmos, first edition, section 9, page 38. Remark on notation

$$\text{Let \{E_{n}\} be a sequence, then}\\$$ $$\lim\limits_{n\rightarrow\infty}E_{n} = \bigcap\limits_{n=1}^{\infty}\bigcup\limits_{m\geq n}^{\infty}E_{m} = \bigcup\limits_{n=1}^{\infty}\bigcap\limits_{m\geq n}^{\infty}E_{m}$$ $$\text{i.e. } \lim\limits_{n\rightarrow\infty}E_{n}=\liminf\limits_{n\rightarrow\infty}E_{n}=\limsup\limits_{n\rightarrow\infty}E_{n}$$

Theorem:

$$\quad$$If $$\mu$$ is a measure on a ring $$\mathcal{R}$$, and if $$\{E_{n}\}$$ is an increasing sequence of sets in $$\mathcal{R}$$ such that $$\lim\limits_{n\rightarrow\infty}E_{n} \in \mathcal{R}$$, then $$\mu(\lim\limits_{n\rightarrow\infty}E_{n}) = \lim\limits_{n\rightarrow\infty}\mu(E_{n})$$

Theorem:

$$\quad$$If $$\mu$$ is a measure on a ring $$\mathcal{R}$$, and if $$\{E_{n}\}$$ is an decreasing sequence of sets in $$\mathcal{R}$$ of which at least one has finite measure and for which $$\lim\limits_{n\rightarrow\infty}E_{n} \in \mathcal{R}$$, then $$\mu(\lim\limits_{n\rightarrow\infty}E_{n}) = \lim\limits_{n\rightarrow\infty}\mu(E_{n})$$

• Thanks for your nice answer. Is there any condition about an uncountable family of $\{E_\alpha\}_{\alpha\in I}$? Commented Jan 15, 2023 at 12:46